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I am having trouble deriving the time dilation. I am using $(-, +, +, +)$ sign convention.

For Minkowski metric, the line element is equal to: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

For a motionless ($dx = dy = dz = 0$) distant observer the line element is equal to: $$ds^2 = -c^2dt^2$$

And for a moving object the line element is equal to: $$ds^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ Where $\tau$ is its proper time.

Now, as far as I understand, I need to set these two line elements equal and solve for $\frac{dt}{d\tau}$: $$-c^2dt^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ $$\left(\frac{dt}{d\tau}\right)^2 = 1 - \frac{\left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2}{c^2}$$ $$\frac{dt}{d\tau} = \sqrt{1 - \frac{v^2}{c^2}}$$

This does not make sense to me, since the time measured by the motionless stationary observer $t$ should be greater than the proper time $\tau$. So $\frac{dt}{d\tau} > 1$. This is not true for $\frac{dt}{d\tau} = \sqrt{1 - \frac{v^2}{c^2}}$. I already saw $\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ somewhere, but I'm not sure how to arrive at it.

For Schwarzschild's metric, I am solving this for $\frac{dt}{d\tau}$: $$-c^2 dt^2 = -\left(1 - \frac{r_s}{r}\right)c^2 d\tau^2 + \left(1 - \frac{r_2}{r}\right)^{-1} dr^2 + r^{2} \left(d\theta^2 + \sin^2 \theta \, d\phi^2\right)$$

Doing this, I arrive at: $$\left(\frac{dt}{d\tau}\right)^2 = \left(1 - \frac{r_s}{r}\right) - \left(1 - \frac{r_2}{r}\right)^{-1} \frac{1}{c^2} \left(\frac{dr}{d\tau}\right)^2 - \left(\frac{r}{c}\right)^{2} \left(\left(\frac{d\theta}{d\tau}\right)^2 + \sin^2 \theta \, \left(\frac{d\phi}{d\tau}\right)^2\right)$$ This doesn't seem even close to being correct. I already saw other posts and the equation is very different. I want to have it universal, so I can plug in any values. Where am I making a mistake? Maybe the change in position over a change in time (e.g. $\frac{dx}{d\tau}$) should be with respect to $t$ instead ($\frac{dx}{dt})$?

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3 Answers 3

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For Minkowski metric, the line element is equal to $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2. \tag{1}$$ In a rest coordinate system ($dx = dy = dz = 0$) time $t$ is equal to proper time $\tau$ $$ds^2 = -c^2d\tau^2. \tag{2}$$ Equating equations $(1)$ and $(2)$ results in $$-c^2 d\tau^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2, \tag{2}$$ $$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{1}{c^2}\left(\Big(\frac{dx}{dt}\Big)^2 + \Big(\frac{dy}{dt}\Big)^2 + \Big(\frac{dz}{dt}\Big)^2\right), \tag{3}$$ and finally in correct equation for the dilatation time $$\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}. \tag{4}$$

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  • $\begingroup$ I think I'm missing something. Is $t$ the time of the motionless clock and $\tau$ the proper time of the moving clock? Or is it the opposite? What are the general steps for any metric? $\endgroup$
    – Yachim
    Nov 3, 2023 at 17:13
  • $\begingroup$ @Yachim A clock in a rest frame measures proper time. $t$ is the so-called coordinate time from the right side of eq. (1). Moving is a relative notion. My clock rests other clock moves in reference to me. From the viewpoint of other clock you can say the same. About general steps for any metric I cannot answer you yet. I have to think about first. $\endgroup$
    – JanG
    Nov 3, 2023 at 19:22
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    $\begingroup$ The sign of $1/c^2$ in $(3)$ should be positive (or $t$ and $\tau$ should swap places). $\endgroup$ Nov 4, 2023 at 1:14
  • $\begingroup$ @KrisWalker You are right! $\endgroup$
    – JanG
    Nov 4, 2023 at 7:37
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    $\begingroup$ @Yachim I have had to correct my equation (3), sorry! Now, you have three correct answers. $\endgroup$
    – JanG
    Nov 4, 2023 at 8:03
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The problem appears to be in your initial setup.

For Minkowski metric, the line element is equal to: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

That is correct.

For a motionless ($dx = dy = dz = 0$) distant observer the line element is equal to: $$ds^2 = -c^2dt^2$$

That is correct but not relevant. The time dilation formula is not restricted to distant observers. Time dilation is the ratio between coordinate time and proper time for a given clock.

And for a moving object the line element is equal to: $$ds^2 = -c^2d\tau^2 + dx^2 + dy^2 + dz^2$$ Where $\tau$ is its proper time.

This one is incorrect. The correct expression is $$ds^2 = -c^2d\tau^2 $$

Deriving time dilation is straightforward. You simply start with the line element, substitute this corrected expression, divide both sides by $-c^2 dt^2$ and simplify. The result is $1/\gamma^2$, where $\gamma$ is the time dilation factor.

$$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$$$-c^2d\tau^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$$$\frac{d\tau^2 }{dt^2} = 1 -\frac{ dx^2 }{c^2 dt^2} -\frac{ dy^2 }{c^2 dt^2}-\frac{ dz^2 }{c^2 dt^2} $$$$\frac{1}{\gamma^2} =1-\frac{v^2}{c^2}$$

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Consider a frame in which the coordinates are $(t, x, y, z)$. An object moving in this frame will have some trajectory with line element $${\rm d}s^2=-c^2{\rm d}t^2+{\rm d}x^2+{\rm d}y^2+{\rm d}z^2.\tag*{(1)}$$ Now boost to the frame of the object, in which the coordinates are $(t^\prime,x^\prime,y^\prime,z^\prime)$. Since the object is at rest in this frame, we have that ${\rm d}x^\prime={\rm d}y^\prime={\rm d}z^\prime=0$, making the line element $${\rm d}s^2=-c^2{\rm d}{t^\prime}^2.\tag*{(2)}$$ This is an invariant, so we recognize $t^\prime$ as an invariant "proper time" of the object and typically denote it as $\tau$, so that ${\rm d}s^2=-c^2{\rm d}\tau^2$. And since it is an invariant, $(1)$ and $(2)$ are equal, so we have that $$-c^2{\rm d}\tau^2=-c^2{\rm d}t^2+{\rm d}x^2+{\rm d}y^2+{\rm d}z^2,\tag*{(3)}$$ which is easily rearranged to obtain ${\rm d}t/{\rm d}\tau=1/\sqrt{1-\boldsymbol{v}\cdot\boldsymbol{v}/c^2}$, where $\boldsymbol{v}={\rm d}\boldsymbol{x}/{\rm d}t$.

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